Question

If $y=\sqrt{(a-x)(x-b)}-(a-b)ta{n}^{-1}\sqrt{\left(\frac{a-x}{x-b}\right),}$ then $\frac{dy}{dx}$ is equal to

• A. $\sqrt{(a-x)(x-b)}$
• B. $\frac{1}{\sqrt{(a-x)(x-b)}}$
• C. $\sqrt{\left(\frac{a-x}{x-b}\right)}$
• D. $\sqrt{\left(\frac{x-b}{a-x}\right)}$

Answer 1

The correct option is C $\sqrt{\left(\frac{a-x}{x-b}\right)}$

Let $x=aco{s}^2\theta +bsi{n}^2\theta$
$\therefore a-x=(a-b)si{n}^2\theta ,x-b=(a-b)co{s}^2\theta \therefore y=(a+b)sin\theta cos\theta -(a-b)\theta =\frac{(a-b)sin2\theta }{2}-(a-b)\theta \Rightarrow \frac{dy}{d\theta }=(a-b)cos2\theta -(a-b)=-(a-b)2si{n}^2\theta =-2(a-b)si{n}^2\theta$
and $\frac{dx}{d\theta }=(b-a)sin2\theta \therefore \frac{dy}{dx}=\frac{-2(a-b)si{n}^2\theta }{(b-a)sin2\theta }=tan\theta =\sqrt{\left(\frac{a-x}{x-b}\right)}$